31 March 2012

Shell Script to swap two numbers without using third variable

By Sandeep 11:13 PM , , , , 3 comments

Q. How do I swap two numbers using addition and subtraction in Bash


Ans:

#!/bin/bash
echo "enter first number"
read a
echo "enter second number"
read b
echo "a before swapping is $a and b is $b"
#swapping
a=$((a+b))
b=$((a - b))
a=$((a-b))
echo "a after swapping is  $a and b is $b"

Output


swapping two numbers using addition and subtraction

Enjoy :)


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3 comments:

  1. AnonymousApril 18, 2013 at 3:15 PM

    Even simpler and without third variable

    #!/bin/bash
    echo "enter first number"
    read a
    echo "enter second number"
    read b
    echo "a before swapping is $a and b is $b"
    #swapping
    read a b <<<"$b $a"
    echo "a after swapping is $a and b is $b"

    ReplyDelete
    Replies
    1. SandeepApril 18, 2013 at 8:43 PM

      nice :)

      Delete
    2. UnknownDecember 17, 2014 at 7:24 AM

      echo "enter the value in var1"
      read n
      echo "entered numbr in var 1 is $n"
      echo "enter the value in var2"
      read m
      echo "entered num in var 2 is $m"
      echo "b4 swapping the values in var1 and var2 are $n and $m"
      var1=$m
      var2=$n
      echo "aftr swapping the values in var1 and var2 are $m and $n"

      Delete